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Solutions Manual for: Communications Systems, 5th edition by Karl Wiklund, McMaster University, Hamilton, Canada Michael Moher, Space-Time DSP Ottawa, Canada and Simon Haykin, McMaster University, Hamilton, Canada Published by Wiley, 2009.
2.14 b)Given that the two energy densities are equal, we only need to prove the result for one. From before, it was shown that the Fourier transform of the half-cosine pulse was:
2.24. a) By the duality and frequency-shifting properties, the impulse response of an ideal low-pass filter is a phase-shifted sinc pulse. The resulting filter is non-causal and therefore not realizable in practice. c)Refer to the appropriate graphs for a pictorial representation. i)Δt=T/100 BT Overshoot (%) Ripple Period 5 9,98 1/5 10 9.13 1/10 20 9.71 1/20 100 100 No visible ripple
2.24 (d) Δt Overshoot (%) Ripple Period T/100 100 No visible ripple. T/150 16.54 1/100 T/200 ~0 No visible ripple. Discussion Increasing B, which also increases the filter’s bandwidth, allows for more of the high-frequency components to be accounted for. These high-frequency components are responsible for producing the sharper edges. However, this accuracy also depends on the sampling rate being high enough to include the higher frequencies.
2.25 BT Overshoot (%) Ripple Period 5 8.73 1/5 10 8.8 1/10 20 9.8 1/20 100 100 - The overshoot figures better for the raised cosine pulse that for the square pulse. This is likely because a somewhat greater percentage of the pulse’s energy is concentrated at lower frequencies, and so a greater percentage is within the bandwidth of the filter.
2.26 b) If B is left fixed, at B=1, and only T is varied, the results are as follows BT Max. Amplitude 5 1.194 2 1.23 1 1.34 0.5 0.612 0.45 0.286 As the centre frequency of the square wave increases, so does the bandwidth of the signal (and its own bandwidth shifts its centre as well). This means that the filter passes less of the signal’s energy, since more of it will lie outside of the pass band. This results in greater overshoot. However, as the frequency of the pulse train continues to increase, the centre frequency is no longer in the pass band, and the resulting output will also be attenuated.
c) BT Max. Amplitude 5 1.18 2 1.20 1 1.27 0.5 0.62 0.45 0.042 Extending the length of the filter’s impulse response has allowed it to better approximate the ideal filter in that there is less ripple. However, this does not extend the bandwidth of the filter, so the reduction in overshoot is minimal. The dramatic change in the last entry (BT=0.45) can be accounted for by the reduction in ripple.
2.27 a)At fs = 4000 and fs = 8000, there is a muffled quality to the signals. This improves with higher sampling rates. Lower sampling rates throw away more of the signal’s high frequencies, which results in a lower quality approximation. b)Speech suffers from less “muffling” than do other forms of music. This is because a greater percentage of the signal energy is concentrated at low frequencies. Musical instruments create notes that have significant energy in frequencies beyond the human vocal range. This is particularly true of instruments whose notes have sharp attack times.
The output will have spectral components at: fm fc fc+ fm fc- fm 2fc 2fm 2fc- fm 2fc+ fm fc- 2fm fc+2 fm 3fc 3fm (c) The bandpass filter must be symmetric and centred around fc . It must pass components at fc+ fm, but reject those at fc+2 fm and higher.
3.10. The circuit can be rearranged as follows: (a)
(b)
Let the voltage Vb-Vd be the voltage across the output resistor, with Vb and Vd being the voltages at each node. Using the voltage divider rule for condition (a):
, , = f b fbb d b d
f b f b f b
R R RRV V V V V V VR R R R R R
−= = −
+ + +
and for (b):
, , =f b fbb d b d
f b f b f b
R R RRV V V V V V VR R R R R R
− += − = − −
+ + +
∴The two voltages are of the same magnitude, but are of the opposite sign.
The transmitted SSB signal is: ˆ[ ( ) cos(2 ) ( )sin(2 )2
cc c
A m t f t m t f tπ π−
Demodulation is accomplished using a product modulator and multiplying by: ' 'cos(2 )c cA f tπ
(a)
' '1 ˆ( ) cos(2 )[ ( )cos(2 ) ( )cos(2 )]2o c c c c cv t A A f t m t f t m t f tπ π π= −
The only lowpass components will be those that are functions of only t and Δf. Higher frequency terms will be filtered out, and so can be ignored for the purposes of determining the output of the detector.
'1 ˆ( ) [ ( ) cos(2 ) ( )sin(2 )]4o c cv t A A m t f t m t f tπ π∴ = Δ − Δ by using basic trig identities.
When the upper side-band is transmitted, and Δf>0, the frequencies are shifted inwards by Δf.
( ) contains {99.98,199.98,399.98} HzoV f∴ (b) When the lower side-band is transmitted, and Δf>0, then the baseband frequencies are shifted outwards by Δf.
This system essentially produces a DSB-SC signal centred around the frequency of y1(t). The lowest frequencies that can be produced are:
1 2 1 2
1 1 2
2 1 2
1( ) [cos(2 ( ) ) cos(2 ( ) )]2
1 MHz 0.9 MHz100 kHz 1.1 MHz
oy t f f t f f t
f f ff f f
π π= − + +
= − == + =
The highest frequencies that can be produced are:
1 1 2
2 1 2
9 MHz 8.1 MHz900 kHz 9.9 MHz
f f ff f f= − == + =
The resolution of the system is the bandwidth of the output signal. Assuming that no branch can be zeroed, the narrowest resolution occurs with a modulation frequency of 100 kHz. The widest bandwidth occurs when there is a modulation frequency of 900 kHz.
3.24 Given the presence of the filters, only the baseband signals need to be considered. All of the other product components can be discarded. (a) Given the sum of the modulated carrier waves, the individual message signals are extracted by multiplying the signal with the required carrier. For m1(t), this results in the conditions:
(b) Given that the maximum bandwidth of mi(t) is W, then the separation between fa and fb must be | fa- fb|>2W in order to account for the modulated components corresponding to fa- fb.
3.25 b) The charging time constant is ( ) 1f sr R C sμ+ = The period of the carrier wave is 1/fc = 50 μs. The period of the modulating wave is 1/fm = 0.025 s. ∴The time constant is much shorter than the modulating wave and therefore should track the message signal very well. The discharge time constant is: 100lR C sμ= . This is twice the period of the carrier wave, and should provide some smoothing capability. From a maximum voltage of V0, the voltage Vc across the capacitor after time t = Ts is:
0 exp( )sc
l
TV VR C
= −
Using a Taylor series expansion and retaining only the linear terms, will result in the
linear approximation of 0 (1 )sC
l
TV VR C
= − . Using this approximation, the voltage will
decay by a factor of 0.94 from its initial value after a period of Ts seconds. From the code, it can be seen that the voltage decay is close to this figure. However, it is somewhat slower than what was calculated using the linear approximation. In a real circuit, it would also be expected that the decay would be slower, as the voltage does not simply turn off, but rather decreases over time.
Problem 3.25. MATLAB code function [y,t,Vc,Vo]=AM_wave(fc,fm,mi) %Problem 3.25 %Inputs: fc Carrier Frequency % fm Modulation Frequency % mi modulation index %Problem 3.25 (a) fs=160000; %sampling rate deltaT=1/fs; %sampling period t=linspace(0,.1,.1/deltaT); %Create the list of time periods y=(1+mi*cos(2*pi*fm*t)).*cos(2*pi*fc*t); %Create the AM wave %Problem 3.25 (b) %%%%Create the envelope detector%%%% Vc=zeros(1,length(y)); Vc(1)=0; %inital voltage for k=2:length(y) if (y(k)>(Vc(k-1))) Vc(k)=y(k); else
Vc(k)=Vc(k-1)-0.023*Vc(k-1); end end %Problem 3.25 (c) %%%Implement the high pass filter%%% %%This implements bias removal Vo=zeros(1,length(y)); Vo(1)=0; RC=.001; beta=RC/(RC+deltaT); for k=2:length(y) Vo(k)=beta*Vo(k-1)+beta*(Vc(k)-Vc(k-1)); end
4.17. Consider the slope circuit response: The response of |X1(f)| after the resonant peak is the same as for a single pole low-pass filter. From a table of Bode plots, the following gain response can be obtained:
1 2
1| ( ) |
1 B
X ff f
B
=−⎛ ⎞+ ⎜ ⎟
⎝ ⎠
Where fB is the frequency of the resonant peak, and B is the bandwidth. For the slope circuit, B is the filter’s bandwidth or cutoff frequency. For convenience, we can shift the filter to the origin (with 1( )X f as the shifted version).
Because the filters are symmetric about the central frequency, the contribution of the second filter is identical. Adding the filter responses results in the slope at the central frequency being:
32 2
| ( ) | 2
(1 )f kB
d X f kdf B k=
= −+
In the original definition of the slope filter, the responses are multiplied by -1, so do this here. This results in a total slope of:
32 2
2
(1 )
k
B k+
As can be seen from the following plot, the linear approximation is very accurate between the two resonant peaks. For this plot B = 500, f1=-750, and f2=750.
Problem 4.24 The amplitude spectrum corresponding to the Gaussian pulse 2 2( ) exp * [ / ]p t c c t rect t Tπ⎡ ⎤= −⎣ ⎦ is given by the magnitude of its Fourier transform.
( ) ( ) ( )
[ ]
2 2
2 2
exp /
exp sinc
P f c c t rect t T
c f c T fT
π
π
⎡ ⎤ ⎡ ⎤= − ⎣ ⎦⎣ ⎦
⎡ ⎤= −⎣ ⎦
F F
where we have used the convolution theorem Problem 4.25 The Carson rule bandwidth for GSM is ( )2TB f W= Δ + where the peak deviation is given by
1 2 / log(2) 0.752 4
fk cf B Bπ
πΔ = = =
With BT = 0.3 and T = 3.77 microseconds, the peak deviation is 59.7 kHz From Figure 4.22, the one-sided 3-dB bandwidth of the modulating signal is approximately 50 kHz. Combining these two results, the Carson rule bandwidth is
( )2 59.7 50
219.4 kHzTB = +
=
The 1-percent FM bandwidth is given by Figure 4.9 with 59.7 1.1950
Beta # of side frequencies 1 1 2 2 5 8 10 14 b)By experimentation, a modulation index of 2.408, will force the amplitude of the carrier to be about zero. This corresponds to the first root of J0(β), as predicted by the theory.
Problem 4.27. a)Using the original MATLAB script, the rms phase error is 6.15 % b)Using the plot provided, the rms phase error is 19.83% Problem 4.28 a)The output of the detected signal is multiplied by -1. This results from the fact that m(t)=cos(t) is integrated twice. Once to form the transmitted signal and once by the envelope detector. In addition, the signal also has a DC offset, which results from the action of the envelope detector. The change in amplitude is the result of the modulation process and filters used in detection.
The above signal has been multiplied by a constant gain factor in order to highlight the differences with the original message signal. c)The earliest signs of distortion start to appear above about fm =4.0 kHz. As the message frequency may no longer lie wholly within the bandwidth of either the differentiator or the low-pass filter. This results in the potential loss of high-frequency message components.
4.29. By tracing the individual steps of the MATLAB algorithm, it can be seen that the resulting sequence is the same as for the 2nd order PLL.
( ) is the phase error ( ) in the theoretical model.ee t tφ The theoretical model of the VCO is:
20
( ) 2 ( )t
vt k v t dtφ π= ∫
and the discrete-time model is: VCOState VCOState 2 ( 1)v sk t Tπ= + − which approximates the integrator of the theoretical model. The loop filter is a PI-controller, and has the transfer function:
( ) 1 aH fjf
= +
This is simply a combination of a sum plus an integrator, which is also present in the MATLAB code:
c)The phase error increases, and tracks the message signal.
d)For a single sinusoid, the track is lost if 0 0 where m f v c vf K K k k A A≥ = For this question, K0=100 kHz, but tracking degrades noticeably around 60-70 kHz. e)No useful signal can be extracted. By multiplying s(t) and r(t), we get:
sin( VCOState) sin(4 VCOState)2c v
f c fA A k f t kφ π φ⎡ ⎤− + + +⎣ ⎦
This is substantially different from the original error signal, and cannot be seen as an adequate approximation. Of particular interest is the fact that this equation is substantially more sensitive to changes in φ than the previous one owing to the presence of the gain factor kv
and 2 2exp( ) exp( )t fπ π− − , then by applying the time-shifting and scaling properties:
2 2 2 2
2
1( ) 2 exp( ( 2 ) )exp( 2 )2
x x x
x
F f f j fπσ π πσ π π μπσ
= −
= 2 2 2exp( 2 2 ) and let 2x xf j f fπ σ μ π ν π− + =
= 2 21exp( )2x xjνμ ν σ−
(b)The value of μx does not affect the moment, as its influence is removed. Use the Taylor series approximation of φx(x), given μx = 0.
2 2
2
0
1( ) exp( )2
exp( )!
x x
n
xxn
φ ν ν σ
∞
=
= −
=∑
0
2 2
0
( )[ ]
1 ( )2 !
nn x
nv
k k kx
xk
dE Xd
k
φ νν
σ νφ ν
=
∞
=
=
⎛ ⎞∴ = −⎜ ⎟⎝ ⎠
∑
For any odd value of n, taking ( )nx
n
ddφ νν
leaves the lowest non-zero derivative as ν2k-n.
When this derivative is evaluated for v=0, then [ ]nE X =0. For even values of n, only the terms in the resulting derivative that correspond to ν2k-n = ν0 are non-zero. In other words, only the even terms in the sum that correspond to k = n/2 are retained.
5.2. (a) All the inputs for x ≤0 are mapped to y = 0. However, the probability that x > 0 is unchanged. Therefore the probability density of x ≤0 must be concentrated at y=0.
(b) Recall that ) 1 where ( ) is an even function.x xf x dx f x∞
−∞
=∫ Because fy(y) is a
probability distribution, its integral must also equal 1.
0 0
( ) 0.5 and ( ) 0.5x yf x dx f y dy+
∞ ∞
∴ = =∫ ∫
Therefore, the integral over the delta function must be 0.5. This means that the factor k must also be 0.5.
and the Paley-Wiener criterion for causality is: 2
( )1 (2 )
fdf
fαπ
∞
−∞
< ∞+∫
For the filter of part (b)
[ ]01( ) ln(2) ln( ( ) ln( )2 xf S f Nα = + −
The first and the last terms have no impact on the absolute integrability of the previous expression, and so do not matter as far as evaluating the above criterion. This leaves the only condition:
The Doppler shift of the frequency observed at the receiver is cD
f vfc
= .
(b) The expectation is given by
( ) ( )
( )
( )0
1exp 2 exp 2 cos2
1 exp 2 sin2
2
n D n n
D n n
D
j f j f d
j f d
J f
π
π
π
π
π τ π τ ψ ψπ
π τ ψ ψπ
π τ
−
−
⎡ ⎤ =⎣ ⎦
=
=
∫
∫
E
where the second line comes from the symmetry of cos and sin under a -π/2 translation.
Eq. (5.174) follows directly from this upon noting that, since the expectation result is real-valued, the right-hand side of Eq.(5.173) is equal to its conjugate.
Problem 5.34 The histogram has been plotted for 100 bins. Larger numbers of bins result in larger errors, as the effects of averaging are reduced. Distance Relative Error 0σ 0.94% 1σ 2.6 % 2σ 4.8 % 3σ 47.4% 4σ 60.7% The error increases further out from the centre. It is also important to note that the random numbers generated by this MATLAB procedure can never be greater than 5. This is very different from the Gaussian distribution, for which there is a non-zero probability for any real number.
5.34 Code Listing %Problem 5.34 %Set the number of samples to be 20,000 N=20000 M=100; Z=zeros(1,20000); for i=1:N for j=1:5 Z(i)=Z(i)+2*(rand(1)-0.5); end end sigma=sqrt(var(Z-mean(Z))); %Calculate a histogram of Z [X,C]=hist(Z,M); l=linspace(C(1),C(M),M); %Create a gaussian function with the same variance as Z G=1/(sqrt(2*pi*sigma^2))*exp(-(l.^2)/(2*sigma^2)); delta2=abs(l(1)-l(2)); X=X/(20000*delta2);
The theoretical values are: μy = 0 (by inspection). The theoretical value of 2
yσ =5.56. See 5.35 (c) for the calculation. 5.35 (b) From the plots, it can be seen that both the real and imaginary components are approximately Gaussian. In addition, from statistics, the sum of tow zero-mean Gaussian signals is also Gaussian distributed. As a result, the filter output must also be Gaussian.
Problem 6.2 The transfer function of the circuit can be found to be
( ) 122
RH fR j fL
j fCπ
π
=+ +
where 12cf LCπ
= and 1 LQR C
=
1( )
1 [( / ) ( / )]c c
H fjQ f f f f
∴ =+ −
For 1Q , the transfer function may be approximated as follows:
1 01 2 ( ) /
( )1 0
1 2 ( ) /
c c
c c
fj Q f f f
H ff
j Q f f f
⎧ >⎪ + −⎪= ⎨⎪ <⎪ + +⎩
For a noise source with a PSD of N0/2, the PSD of the filtered output will be
02 2 2
02 2 2
/ 2 01 4 ( ) /
( )/ 2 0
1 4 ( ) /
c cN
c c
N fQ f f f
S fN f
Q f f f
⎧ >⎪ + −⎪= ⎨⎪ >⎪ + +⎩
which is a symmetric function about the f = 0 axis. However,
( ) ( ) ( ) ( )NI NQ N c N cS t S t S f f S f f= = − + + Around f = fc, this allows us to approximate the PSDs of the in-phase and quadrature components as follows
The variance of the in-phase and quadrature components of n(t) however, are the same as the variance of n(t) itself. Therefore, by taking the inverse Fourier transform of the above PSD and setting τ=0, we obtain
1( ) exp( )21(0)2
c cN
cN
f fRQ QfR
Q
π πτ τ
π
= −
=
which is the approximate variance (power) of the narrow band noise. Therefore, the SNR is,
Figure 6.16 Plot from Matlab script (a) AM modulated carrier (b) AM modulated carrier plus noise (c) AM demodulated signal in absence of noise (d) AM demodulated signal in noise
Figure 6.17 Plot from Matlab script (a) FM modulated carrier (b) FM modulated carrier plus noise (c) FM demodulated signal in absence of noise (d) FM demodulated signal in noise
(a) The worst case ISI occurs if all preceding pulses have the same polarity
In this case, the received signal is
1
0
( ) ( )
( ) ( )
ii
i
r t a p t iT
a p t p t iT
∞
=−∞
−
=−∞
= −
= + −
∑
∑
where we have used the fact that the pulse is one sided. Substituting the pulse shape in the summation, we obtain
[ ]
1
0
01
( ) ( ) exp
( ) exp exp
i
i
t iTr t a p tT
ta p t iT
−
=−∞
∞
=
−⎡ ⎤= + −⎢ ⎥⎣ ⎦⎡ ⎤= + − −⎢ ⎥⎣ ⎦
∑
∑
Recognizing that this is a geometric summation, we obtain 1
0 1
0
0
( ) ( ) exp1
1( ) exp1
( ) 0.582 ( )
t er t a p tT eta p tT e
a p t p t
−
−⎡ ⎤= + −⎢ ⎥ −⎣ ⎦⎡ ⎤= + −⎢ ⎥ −⎣ ⎦
= +
In this example the ISI is nearly 60% of the original pulse. (b) If the time constant τ is not equal to T, then the calculation of the part (a) can be repeated for the generic case. In which case, we obtain the following expression
( )0
1( ) ( ) expexp / 1
tr t a p tTτ τ
⎡ ⎤= + −⎢ ⎥ −⎣ ⎦
If the maximum reduction of the eye opening is 20%, then by solving
10.20exp( / ) 1T τ
=−
we find that τ = T/ln(6) = 0.558T. Recall from Chapter Example 2.2, that the spectrum of the one-sized exponential pulse is
and thus the 3-dB bandwidth is B3dB = 1/2πτ. By decreasing τ to 55.8% of T, we have increased the 3-dB bandwidth of the signal by the inverse of this amount, in order to keep the ISI to a manageable amount. Problem 8.18 Since the analog frequency response of the system including the matched filter P(f) is given by ( ) exp ( )H f f T P f⎡ ⎤= −⎣ ⎦ (1) the aliased version of this frequency response is given by
( ) expan
n nH f f T P fT T
∞
=−∞
⎡ ⎤ ⎛ ⎞= − + +⎜ ⎟⎢ ⎥ ⎝ ⎠⎣ ⎦∑ for 2f
T< . (2)
For zero ISI we must have Ha(f) = 1 for 12
fT
< . There are many solutions to this
problem. To reduce the number of options, we simply choose a function H(f) that satisfies
Problem 8.24 The provided Matlab script uses an approximation to the telephone model shown in Figure 8.9, but not exactly the same model used in Example 8.2. The response when using a 1.6 kbps NRZ signal is shown below. The response shows the same qualitative response as in Example 8.2 with level droop because the channel does not pass dc.
0 20 40 60 80 100 120 140 160 180 200
-1.5
-1
-0.5
0
0.5
1
1.5
Time samples
Am
plitu
de
Figure 8.24a-1 Response with NRZ signal.
The modified script for simulating the Manchester code is the following: %-------------------------------------------- % Problem 8.24 Telephone channel with Manchester code %-------------------------------------------- Fs = 32; % sample rate (kHz) Rs = 1.6; % symbol rate (kHz) Ns = Fs/Rs; % samples per symbol Nb = 30; % number of bits to simulate %--- Discrete B(z)/A(z) model of telephone channel --- A = [1.00, -2.838, 3.143, -1.709, 0.458, -0.049]; B = 0.1*[1.0, -1.0]; %---------------------------------------------------------------- % Simulate performance %================================================================ % pulse = [ones(1,Ns)]; % bipolar NRZ pulse pulse = [ones(1,Ns/2) -ones(1,Ns/2)]; % Manchester line code data = sign(randn(1,Nb)); Sig = pulse' * data; Sig = Sig(:); %--- Pass signal through channel ---- RxSig = filter(B,A,Sig); %--- Plot results ------------------------ plot(real(RxSig))
hold on, plot(Sig,'r'), hold off xlabel('Time samples'),ylabel('Amplitude') axis([0 200 -1.75 1.75]) % don't plot all samples
The signal output before and after the telephone line with a 1.6 kbps Manchester code is the following:
0 20 40 60 80 100 120 140 160 180 200
-1.5
-1
-0.5
0
0.5
1
1.5
Time samples
Am
plitu
de
Figure 8.24a-2 Response with Manchester code at 1.6 kbps.
The output shows much less level droop because the individual pulses do not have a dc component. If we increase the data rate to 3.2 kbps we obtain the following response:
Figure 8.24a-3 Response with Manchester code at 3.2 kbps.
With the higher data, there is significant distortion of the signal due to the fact that the bandwidth of the signal exceeds that of the channel. (b) To included match filtering and plot the eye diagram, add the lines
after the signal has been passed through the channel model. The eye diagram with the 1.6 kbps Manchester code after matched filtering is shown below. There are two eyes present in the diagram, one at 0.6 and 1.6 symbol offsets. The eye is open to almost the full amplitude of the signal indicating that the signaling format is quite robust in the presence of noise. The eye, however, is quite narrow indicating it is not tolerant of large timing errors.
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-20
-15
-10
-5
0
5
10
15
20
Symbol Periods
Figure 8.24b-1 Eye diagram with Manchester code at 1.6 kbps.
The eye diagram with the 3.2 kbps Manchester code after matched filtering is shown below. There are also two eyes present in the diagram, one at 0.8 and 1.8 symbol offsets. The eye is open, but the opening is significantly less than with the slower transmission rate, indicating a greater susceptibility to noise.
Figure 8.24b-2 Eye diagram with Manchester code at 3.2 kbps.
(c) Implement 4-ary signaling by changing
data = sign(randn(1,Nb)); to
data = 2*floor(4*rand(1,Nb)) - 3; which produces random symbols ±1, and ±3. The eye diagram with a 1.6 kHz symbol rate is shown below. The diagram shows three open eyes clearly separating the four levels and only a small amount of intersymbol interference.
The height of the eye opening is relatively unchanged with α = 0.5 but we find the eye width is slightly less indicating a greater sensitivity to timing errors. (b) The results with τ = T are shown below for α = 1.0 and 0.5 respectively. The larger value of τ implies that the channel has a narrower bandwidth. This is turn causes more ISI, which becomes evident with the reduced eye opening.
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-0.08
-0.06
-0.04
-0.02
0
0.02
0.04
0.06
0.08
Symbol Periods Figure 8.25b-1 Eye diagram with α=1.0 and τ = T.
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-0.08
-0.06
-0.04
-0.02
0
0.02
0.04
0.06
0.08
Symbol Periods Figure 8.25b-2 Eye diagram with α=0.5 and τ = T.
Problem 8.26 We create the following function to compute the equalizer,
%----------------------------------------- % Problem 8.26 - Compute equalizer %------------------------------------------- function w = equalizer(h,pulse,N, Ns); %-- Compute system impulse response --- c = conv(h,pulse); % combine tx pulse shape and channel response [peak,centre] = max(abs(c)); % locate peak of impulse response and define as centre centre = round(centre); %--- Compute "C" matrix for two cases of N=3 or 5 --- if (N==3) C = [c(centre) c(centre-Ns) c(centre-2*Ns)]; C = [C; ... c(centre+Ns) c(centre) c(centre-Ns)]; C = [C; ... c(centre+2*Ns) c(centre+Ns) c(centre)]; b= [ 0 1 0]'; elseif (N==5) % note coefficient is zero if index is 0 or negative C = [c(centre) c(centre-Ns) c(centre-2*Ns) 0 0]; C = [C; ... c(centre+Ns) c(centre) c(centre-Ns) c(centre-2*Ns) 0]; C = [C; ... c(centre+2*Ns) c(centre+Ns) c(centre) c(centre-Ns) c(centre-2*Ns)]; C = [C; ... c(centre+3*Ns) c(centre+2*Ns) c(centre+Ns) c(centre) c(centre-Ns)]; C = [C; ... c(centre+4*Ns) c(centre+3*Ns) c(centre+2*Ns) c(centre+Ns) c(centre)]; b= [ 0 0 1 0 0]'; else ['N not supported'] end %--- Compute equalizer --- w = inv(C)*b; w = w/max(abs(w)); % Normalize equalizer return
To apply an equalizer to the signal of Problem 8.25, we modify the script to the following.
%------------------------------------------------------ % Prob 8.26 Equalized RC pulse shaping %------------------------------------------------------ T = 1; % symbol period Rs = 1/T; % symbol rate Ns = 16; % number of samples per symbol Fs = Rs*Ns; % sample rate (kHz) Nb = 1000; % number of bits to simulate alpha = 0.5; % rolloff of raised cosine N = 3; % number of equalizer taps %--- Discrete model of channel --- t = [0 : 1/Fs : 5*T]; h = exp(-t / (T)) /Fs; % impulse response scaled for sample rate pulse = firrcos(5*Ns, Rs/2, Rs*alpha, Fs); % 100% raised cosine filter %--- compute equalizer ---
w = equalizer(h,pulse,N, Ns); w = [w(:) zeros(N,Fs-1)]'; % upsample to Fs, so we can generate eye diagram w = w(:); %--- Pulse shape the data ----- data = sign(randn(1,Nb)); % random binary data Udata = [1; zeros(Fs-1,1)] * data; % upsample data Udata = Udata(:); % " Sig = filter(pulse,1,Udata); % pulse shape data Sig = Sig((length(pulse)-1)/2:end); % remove filter delay %--- Pass signal through the channel ---- RxSig = filter(h,1,Sig); %--- Equalize signal ---- EqSig = filter(w,1, RxSig); %--- Plot results ------------------------ ploteye(EqSig(4*Fs:end), Fs); % ignore initial transient xlabel('Symbol Periods')
Then with a N=3 tap equalizer, we obtain the following eye diagram. This is a huge improvement over the eye diagram without equalization that we observed in Problem 8.25b. The eye opening increases from 10% of the peak amplitude to approximately 80% of the peak amplitude at the sample instant.
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-0.06
-0.04
-0.02
0
0.02
0.04
0.06
Symbol Periods
Figure 8.26a Eye diagram with α=0.5, τ = T, and N=3 equalizer.
When the number equalizer taps is increased to N=5, the further improvement for this channel is small as seen by the following eye diagram.
Problem 9.1 The three waveforms are shown below for the sequence 0011011001. (b) is ASK, (c) is PSK; and (d) is FSK.
Problem 9.2 The bandpass signal is given by ( )( ) ( ) cos 2 cs t g t f tπ= The corresponding amplitude spectrum, using the multiplication theorem for Fourier transforms, is given by
[ ]( ) ( )* ( ) ( )
( ) ( )c c
c c
S f G f f f f fG f f G f f
δ δ= − + +
= − + +
For a triangular spectrum G(f), the corresponding sketch is shown below. Problem 9.3 (a) From Example 5.12, we can compute the bandpass spectrum directly from the spectrum of the baseband equivalent representation. The baseband representation of the signal is
( ) ( ) ( )I Qg t g t jg t= + The autocorrelation of this complex random process is ( ) ( ) ( ) ( ) ( )g I Q IQ QIR R R jR jRτ τ τ τ τ= + − + The corresponding baseband power spectrum is then ( ) ( ) ( ) ( ) ( )I Q QI IQS f G f G f j G f G f⎡ ⎤= + + −⎣ ⎦ If the two random processes are independent and zero mean, then the cross-correlations are zero (and so are the corresponding cross-spectra). Then, the baseband spectrum is given by ( ) ( ) ( )I QG f G f G f= + The corresponding bandpass spectrum is then
(b) If gI(t) and gQ(t) are independent NRZ line codes then the corresponding baseband power spectra are 2( ) ( ) sinc( )I Q b bG f G f A T fT= = So bandpass spectrum is
( ) ( )2 214( ) sinc ( ) sinc ( )b c b b c bS f A T f f T A T f f T⎡ ⎤= − + +⎣ ⎦
And the spectrum looks like the following
Figure 9.3b. Bandpass spectrum with baseband NRZ line codes.
(c) If gI(t) = -gQ(t) then the signals are not independent, and the cross-spectral densities are not zero. However in this case ( ) ( )IQ QIR Rτ τ= and we obtain the same result, as if they were independent. (d) In this case where pulse shaping has a raised cosine spectral shape, the bandpass signal has the spectrum shown in the following.
Problem 9.13 Problem 9.13 A block diagram of a simple non-coherent detector consists of an energy integrator as shown in Figure 9.13(a). Note that the detector must be sampled and then cleared at the end of each bit interval
s(t)ξ Compare
toThreshold
2
0
( )bT
s t dt∫
EnergyDetector
Figure 9.13a Non-coherent detector.
A block diagram of a more complex coherent detector for ASK consists of a coherent down conversion to baseband followed by an integrate and dump detector. As with the energy detector of part (a), the integrate and dump detector must be sampled and then cleared at the end of each bit interval. The additional complexity with the coherent detector arises in the need for a carrier recovery circuit, and we are compensated for this additional complexity with better performance.
Problem 9.19 The important point to note here, in comparison to the plotted results, is that the error performance of the coherent QPSK is slightly degraded with respect to that of coherent PSK and coherent MSK. Otherwise, the observations made in Section 9.5 still hold here.
As described in Problem 9.3, a random sequence using this pulse shape will have a spectrum 2( ) ( )S f H f= .Comparing this to the sinc function which corresponds to the spectrum of a rectangular pulse, as shown in part (a), we find that the tails of the power spectrum decrease much faster, according to f4, with the MSK pulse shape.
Problem 9.23 (a) Let xI0 and xQ0 denote the in-phase and quadrature components of the matched filter output in the lower path of Figure 9.12 when a “1” is transmitted. Then the output of the enveloped detector is given by 2 2
0 0 0I Ql x x= + (1) Now the channel noise is w(t) is both white with power spectral density N0/2 and Gaussian with zero mean. Correspondingly, we find that the random variables XI0 and XQ0 (represented by samples xI0 and xQ0) are both Gaussian-distributed with zero mean and variance N0/2, given the phase θ. Hence we may write
0
20
000
1( ) expI
IX I
xf xNNπ
⎛ ⎞= −⎜ ⎟
⎝ ⎠ (2)
and
0
20
000
1( ) expQ
QX Q
xf x
NNπ
⎛ ⎞= −⎜ ⎟⎜ ⎟
⎝ ⎠ (3)
Note that XI0 and XQ0 are independent Gaussian random variables, and so we may express there joint probability density function by
( )0 0
2 20 0
, 0 00 0
1, expI Q
I QX X I Q
x xf x x
N Nπ⎛ ⎞+
= −⎜ ⎟⎜ ⎟⎝ ⎠
(4)
Recall the rectangular-to-polar conversion 0 0 cosIx l θ= (5) 0 0 sinQx l θ= (6) In a limiting sense, we may equate the two areas of the different co-ordinate systems 0 0 0 0I Qdx dx l dl dθ= (7) where l0 and θ are the envelope and phase of the observed process. Then substituting (5) and (6) into (4), we find that the probability of the random variables L0 and Θ lying in the area defined by (7) is
2
0 00
0 0
expl l dl dN N
θπ
⎡ ⎤−⎢ ⎥⎣ ⎦
(8)
and the joint probability density function of L0 and Θis given by
( )0
20 0
, 00 0
, expLl lf lN N
θπΘ
⎡ ⎤= −⎢ ⎥
⎣ ⎦ (9)
This probability function is independent of the angle θ, and consequently L0 and Θ are statistically independent, i.e., ( ) ( ) ( )
0 0, 0 ,L L of l f l fθ θΘ Θ= . In particular, the phase is uniformly distributed inside the range 0 to 2π as shown by
This leaves the probability density function of the random variable L0 as
( )0
20 0
00 0 0
2 exp 0
0L
l l lf l N N
elsewhere
⎧ ⎡ ⎤− ≥⎪ ⎢ ⎥= ⎨ ⎣ ⎦
⎪⎩
(11)
which is the Rayleigh probability density function. (b) The output of the upper envelope detector of Figure 9.12, when a “1” is sent, is the equivalent of a sinusoid plus noise. A sample function of the sinusoidal wave plus noise is then expressed by ( )( ) cos 2 ( )c cx t A f t n tπ= + (12) Representing the narrowband noise n(t) in terms of its in-phase and quadrature components, we may write ( ) ( )'( ) ( ) cos 2 ( )sin 2I c Q cx t n t f t n t f tπ π= + (13) where ' ( ) ( )I In t A n t= + (14) We assume that n(t) is Gaussian with zero mean and variance N0/2. Accordingly, we may state the following:
(i) Both ' ( )In t and ( )Qn t are Gaussian and statistically independent.
(ii) The mean of ' ( )In t is A and that of ( )Qn t is zero.
(iii) The variance of both ' ( )In t and ( )Qn t is N0/2.
We may therefore express the joint probability density function of the random variables 'IN and QN corresponding to ' ( )In t and ( )Qn t as follows;
( )
'
2' 2'
,0 0
1( , ) expI Q
I QI QN N
n A nf n n
N Nπ
⎡ ⎤− +⎢ ⎥= −⎢ ⎥⎣ ⎦
(15)
Let l1 denote the envelope of x(t) and θ denote its phase. From the complex baseband equivalent of Eq. (13) we find that
Following a procedure similar to that described in the derivation of the Rayleigh distribution, we find that the joint probability density function of the random variables L1 and Θ, corresponding to l1 and θ for some fixed time t, is given by
1
2 21 1 1
,0 0
2 cos( , ) expLl l A Alf rN N
θθπΘ
⎡ ⎤+ −= −⎢ ⎥
⎣ ⎦ (18)
We see that in this case, however, we cannot express the joint probability density function , ( , )Rf r θΘ as a product ( ) ( )Rf r f θΘ . This is because we now have a term involving the values of both random variable multiplied together as rcos θ. Hence, L1 and Θ, are dependent. We are interested, in particular, in the probability density function of L1. To determine this probability density function, we integrate Eq. (18) over all possible values of θ, obtaining the marginal density
1 1
2
1 , 10
2 2 21 1 10
0 0 0
( ) ( , )
2 21exp exp cos2
RLf l f l d
l l A Al dN N N
π
π
θ θ
θ θπ
Θ=
⎡ ⎤ ⎛ ⎞+= − ⎜ ⎟⎢ ⎥
⎣ ⎦ ⎝ ⎠
∫
∫ (19)
The integral on the right-hand side of Equation ( ) can be identified in terms of the defining integral for the modified Bessel function of the first kind of zero order (see Appendix), that is
( )2
0 0
1( ) exp cos2
I x x dπ
θ θπ
= ∫ (20)
Thus, letting 21x Al σ= , we may rewrite Eq. (19) in the compact form
1
2 21 1 1
1 00 0 0
2 2( ) expLl l A Alf l I
N N N⎡ ⎤ ⎛ ⎞+
= − ⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠
for 1 0l ≥ (21)
This is the Rician distribution. (c) The probability of an error is the probability that L0 > L1 when a “1” is transmitted and so